# Lecture 7 From NFA to Regular Language Lecture 8 From NFA to Regular Language Induction on k= # of states other than initial and final states K=0 a a* b c*a(d+bc*a)* a c d k>1 ac*e

c a e ac*d bc*e d b bc*d Example 1 0 1 0+10

0 1 0 Why? 10 0 1 1 (0+10)*0(1+10(0+10)*0)* 0+10 1

0 (0+10)*0 How to get in the final state? 1 0+10 0 10 10(1+10)*0 Cycles at the final state. Example 2 0 1 0

0 1 0 0 0 1 0 0 0 1 1

1 0 0 1 1 1 1 1 0 1

Solution 1 (0+10)*0(1+10(0+10)*0)* + 0*(1+01*1)(00*(1+01*1))* 0 0 1 0 0 1 1 Solution 2

0 1+01*1 01* 0+(1+01*1)0 (0+(1+01*1)0)*(01*+1+01*1) 01*+(1+01*1) 0 0+10 1 1 0

0 1 1 +1 10 1 Solution 3 (0+10+01*10)*(1+01*(+1)) 0+10+01*10 1+01*(+1) Different ways may give different regular expressions for the same language.

Theorem A language is regular if and only if it can be accepted by an NFA if and only if it can be accepted by a DFA. Closure Properties If A is regular, so is its complement A. If A and B are regular, then AB, A\B, A U B, AB are regular. Quotient L1/L2 = {x | there exists y in L2 s.t. xy in L1} If L1 is regular, so is L1/L2 . L1=L(M), M=(Q,,,s,F) L1/L2 = L(M), M = (Q,,,s, F) where F={q in Q | there exists y in L2 s.t. (q,y) in F} Example L2 = 0*1+0*11

L1 = L(M) 0 0 1 1 0 1 1 0